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A man can view through telescope two district objects at a distance of $120 \;m$ from him if he uses light of $ \lambda= 550 A^{\circ}$ What must be minimum separation between objects if diameter of his pupil $D= 2.5 \;mm$

$(a)\;0.032\;m \\ (b)\;1.02\;m \\ (c)\;0.06\;m \\ (d)\;2.10\;m $

1 Answer

Limit of resolution for small angle is given by
$ \theta_{min} =\large\frac{1.22 \lambda}{D}$
$\qquad= \large\frac{5550 \times 10^{-10}}{2.5 \times 10^{-3}} \times$$ 1.22$
$\qquad= 2.7 \times 10^{-4}$
Separation between objects
$ \qquad= \theta_{min} \times distance $
$\qquad= 2.7 \times 10^{-4} \times 120$
$\qquad= 0.032\;m$
Hence a is the correct answer.
answered Jan 21, 2014 by meena.p

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